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ira [324]
4 years ago
8

A solution contains 15 ppm of benzene. The density of the solution is 1.00 g/mL. This means that ________.

Chemistry
1 answer:
snow_tiger [21]4 years ago
6 0

Answer:

  • second choice:<em><u> 1.0 g of the solution contains 15 × 10⁻⁶ g of benzene.</u></em>

Explanation:

ppm is a unit of concentration that means parts per million. In grams that is grams of solute per one million (10⁶) grams of solution.

Then, <em>15 ppm of benzene</em> means that there are 15 grams of benzen in 1,000,000 grams of solution.

That leads to:

  • 1,000,000 g solution / 15 g benzene

Multiplying numerator and denominator by 10⁻⁶ you find:

  • 1,000,000 × 10⁻⁶ g solution / (15 × 10⁻⁶ g benzene)

Simplifying:

  • 1.0 g solution / (15 × 10⁻⁶ g benzene)

Which is read as 1.0 g of the solution contains 15 × 10⁻⁶ g of benzene, i.e. the second answer choice.

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A 7.300 gram sample of aluminum was quantitatively combined with a sample of selenium to form a compound. The compound weighed 3
adelina 88 [10]

Answer:

The empirical formula is Al2Se3  

Explanation:

<u>Step 1:</u> Data given

Mass of the sample of aluminium = 7.300 grams

Mass of the compound = 39.35 grams

Molar mass of aluminium = 26.98 g/mol

Molar mass of selenium = 78.96 g/mol

<u>Step 2:</u> Calculate mass of selenium

Mass of selenium = mass of compound - mass of aluminium

Mass of selenium = 39.35 - 7.3 = 32.05 grams

<u> Step 3:</u> Calculate moles of Al

Moles Al = Mass Al/ Molar mass Al

Moles Al = 7.300 grams / 26.98 g/mol

Moles Al = 0.2706 moles

<u>Step 4:</u> Calculate moles of Se

Moles Se = Mass Se / Molar mass Se

Moles Se =32.05 g / 78.96 g/mol

Moles Se = 0.4059 moles

<u>Step 5</u>: Divide through the smallest amount of mol

Aluminium: 0.2706 / 0.2706 = 1

Selenium: 0.4059/0.2706 = 1.5

This means for each mol aluminium, we have 1.5 moles of selenium

For each 2 moles of aluminium, we have 3 moles of selenium

The empirical formula is Al2Se3  

This is aluminium(III) selenide.

8 0
3 years ago
a compound contains 69.94 percent iron and 30.06 percent oxygen. what is its molecular formula if the molar mass of the compound
anyanavicka [17]
The percentages are given in weight. You need to transform them into moles.

Take a basis of 100 grams of compound. Then there are 69.94 grams of iron and 30.06 grams of oxygen.

Use the atomic masses of iron and oxygen to obtain each number of moles.

Molar mass of iron: 55.85 g/mol
# moles of iron = 69.94 g / 55.85 g/mol = 1.252

Molar mass of oxygen: 16 g/mol
# moles of oxygen = 30.06 g / 16 g/mol =1.87875

Proportions

1.87875  moles of Oxigen/ 1.252moles of iron = 1.50 /1

3moles of Oxygen: 1 mol of Iron

Empirical formula: FeO3

Molar mass of the empirical formula: 55.85 g/mol + 3(16g/mol) = 103.85 g/mol

Molar mass of the compound / molar mass of the empirical formula = 199.55 g/mol / 103.85 g/mol = 1.92

Round to 2.

Then the molecular formula is the empirical formula times 2.

Fe2O6
6 0
3 years ago
Plesse help asap!!!​
Naily [24]

Answer:

4.4×10² cm³

Explanation:

From the question given above, the following data were obtained:

Diameter (d) = 68.3 mm

Height (h) = 0.120 m

Volume (V) =?

Next, we shall convert the diameter (i.e 68.3 mm) to cm.

This can be obtained as follow:

10 mm = 1 cm

Therefore

68.3 mm = 68.3 mm / 10 mm × 1 cm

68.3 mm = 6.83 cm

Therefore, the diameter 68.3 mm is equivalent 6.83 cm.

Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:

1 m = 100 cm

Therefore,

0.120 m = 0.120 m/ 1 m × 100 cm

0.120 m = 12 cm

Therefore, the height 0.120 m is equivalent 12 cm.

Next, we shall determine the radius of the cylinder. This can be obtained as follow:

Radius (r) is simply half of a diameter i.e

Radius (r) = Diameter (d) /2

r = d/2

Diameter (d) = 6.83 cm

Radius (r) =?

r = d/2

r = 6.83/2

r = 3.415 cm

Finally, we shall determine the volume of the cylinder as follow:

Radius (r) = 3.415 cm

Height (h) = 12 cm

Volume (V) =?

Pi (π) = 3.14

V = πr²h

V = 3.14 × (3.415) ² × 12

V = 440 cm³

V = 4.4×10² cm³

Therefore, the volume of the cylinder is 4.4×10² cm³

8 0
3 years ago
Assuming the diameters of the nucleus and atom are 10A
victus00 [196]

Answer:

2.95 \times 10^{-9} is the fraction of the nucleus is present in atom.

Explanation:

Given:

The diameter of the nucleus - 10A

The diameter of atom - 1.5 A

From this, we can find the radius of nucleus and atom i.e.

For nucleus =\frac{10}{2}

For atom = \frac{1.5}{2}

Now, the fraction of nucleus = \frac{volume of nucleus}{volume of atom}

                                               = \frac{4}{3}\pi (\frac{10}{2})\frac{3.14}{2}\pi (\frac{10}{2})^3

  or,                                         = (\frac{10}{1.5})^3

                                               = 2.95 \times 10^{-9}

Therefore, 2.95 \times 10^{-9} is the fraction of nucleus in the atom.

8 0
3 years ago
BF3 is the Lewis____in the
kakasveta [241]

Answer:

A

Explanation:

Its an acid.

8 0
3 years ago
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